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I'm having trouble figuring out the response of this circuit, specifically the voltage across the capacitor in my homework. The input from the source is a unit step function, and there are no initial conditions for the capacitor or inductor.

schematic

simulate this circuit – Schematic created using CircuitLab

I first turned it into a norton equivalent, and used node analysis to arrive at the differential equation.

enter image description here

$$\frac{d^2}{dt}e_1+\frac{d}{dt}e_1(\frac{R_2}{L}+\frac{1}{R_1C})+e_1(\frac{1}{LC}+\frac{R_2}{CLR_1})=\frac{d}{dt}\frac{R_2}{C}I_1+\frac{R_2}{LC}I_1$$

I know that the solution to this equation takes the form:

$$ e_1(t) = Ae^{-\alpha t}cos(\omega_dt+\phi)$$

So I think my next step is to find A and phi by the following expressions:

$$e_1(0)= ?$$

$$\frac{d}{dt}e_1(0) = ?$$

However, I am a bit unsure about how exactly to go about this. My reasoning suggests that they are both 0 - at the moment the source switches from 0v to 1v, the capacitor is a short circuit and the inductor is an open circuit. So all the current flows through the capacitor, and there is no voltage drop since the current source is shorted - so \$e_1 = 0\$ and \$e_1'=0\$. However, this means that A and phi are also both zero, so this is confusing for me. Can someone give me some help with finding A and phi?

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At the instant of switch on, the capacitor is the dominant impedance and therefore the voltage across it (e1) is zero however it will be rising at a non-zero rate: -

enter image description here

As you can see, the charging voltage rate (dv/dt) equals I/C where I is the current source in your modified circuit (or V1/R1 in your original circuit).

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  • \$\begingroup\$ Yes, I think I see what you mean - let me try this out and I'll get back to you. \$\endgroup\$ – Plasma Nov 15 '16 at 20:21
  • \$\begingroup\$ This did work for me - but I had a general follow-up. To determine the initial conditions, should I replace the inductors with an open and the capacitor with a short and then consider voltage and current through the capacitor? \$\endgroup\$ – Plasma Nov 16 '16 at 4:05
  • \$\begingroup\$ The inductor is effectively an open circuit at start up and for the capacitor it's just the regular shape of the exponential curve right at t(0)+ that is considered. \$\endgroup\$ – Andy aka Nov 16 '16 at 8:24
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If the capacitor and inductor have zero initial energy conditions, then the voltage across the capacitor Vc at t = 0 would be 0V since the capacitor is shorted. Note that this is only true for zero initial energy conditions.

If it does not state that, then you would have to solve for continuity variables (i.e. Vc and iL where iL is the current through the inductor) at t=0. After getting those, solve the circuit for t>0.

Although, does it state that the circuit is in DC steady state at t = 0-? If yes, then it should be helpful in finding Vc(0). If no, are there any statements regarding the ckt or what you told us is just that?

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  • \$\begingroup\$ I forgot to mention, there are no initial conditions in the circuit (post updated to reflect this) \$\endgroup\$ – Plasma Nov 15 '16 at 16:33
  • \$\begingroup\$ You could assume that the circuit has zero initial conditions, and compare your answer to perhaps a solution key? \$\endgroup\$ – user128233 Nov 16 '16 at 0:27
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\$e_1(0)=0\$

\$\frac{d}{dt}e_1(0)=\frac{V1}{R1C1}\$ from \$\frac{dV}{dt}=Ic/C\$ and \$Ic=V1/R1\$ after t=0

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  • \$\begingroup\$ I answered your question on initial conditions. whoever thinks otherwise is incorrect (-1). next question? The step input defines e1(t)=0 for T<0, the capacitor voltage rises as does the inductor current , the next question is the Q , overshoot and resonant frequency of the step response. \$\endgroup\$ – Tony Stewart EE75 Nov 15 '16 at 13:54

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